Question: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $z \neq 0$. $n = \dfrac{9z - 27}{-3z - 27} \times \dfrac{z^2 + 8z - 9}{-6z + 6} $
Answer: First factor the quadratic. $n = \dfrac{9z - 27}{-3z - 27} \times \dfrac{(z + 9)(z - 1)}{-6z + 6} $ Then factor out any other terms. $n = \dfrac{9(z - 3)}{-3(z + 9)} \times \dfrac{(z + 9)(z - 1)}{-6(z - 1)} $ Then multiply the two numerators and multiply the two denominators. $n = \dfrac{ 9(z - 3) \times (z + 9)(z - 1) } { -3(z + 9) \times -6(z - 1) } $ $n = \dfrac{ 9(z - 3)(z + 9)(z - 1)}{ 18(z + 9)(z - 1)} $ Notice that $(z - 1)$ and $(z + 9)$ appear in both the numerator and denominator so we can cancel them. $n = \dfrac{ 9(z - 3)\cancel{(z + 9)}(z - 1)}{ 18\cancel{(z + 9)}(z - 1)} $ We are dividing by $z + 9$ , so $z + 9 \neq 0$ Therefore, $z \neq -9$ $n = \dfrac{ 9(z - 3)\cancel{(z + 9)}\cancel{(z - 1)}}{ 18\cancel{(z + 9)}\cancel{(z - 1)}} $ We are dividing by $z - 1$ , so $z - 1 \neq 0$ Therefore, $z \neq 1$ $n = \dfrac{9(z - 3)}{18} $ $n = \dfrac{z - 3}{2} ; \space z \neq -9 ; \space z \neq 1 $